$x$ is accumulation point of $A\subset X$ iff each neighborhood of $x$ contain infinite point of $A$

Question

Let $X$ be a hausdorff space, let $A\subset X$. Show that $x$ is accumulation point of $A$ iff each neighborhood of $x$ contain infinite point of $A$

I thought to say that if $x$ is accumulation point and $X$ is a hausdorff space then $x\in A$ but who promise that there are infinite points in $A$?

Answer 1

Suppose a neighborhood $U$ of $x$ contains only finitely many points of $A$ distinct from $x$, say $a_1,\dots,a_n$.

Since $X$ is Hausdorff, for every $i=1,2,\dots,n$, there is a neighborhood $U_i$ of $x$ such that $a_i\notin U_i$ (note that just the $T_1$ property is used).

Consider $V=U\cap U_1\cap U_2\cap\dots\cap U_n$. Then $V$ is a neighborhood of $x$ and $V\cap A\subseteq\{x\}$. Therefore $x$ is not an accumulation point for $A$.

Answer 2

$\mathbb{R}$ is separated and $0$ is an accumulation point of $(0,1)$ but $0$ is not an element of $(0,1)$.

Suppose that $X$ is separated, let $x$ be an accumulation point of $A$, suppose that there exists a neighborhood $U$ of $x$ such that $U\cap A=\{a_1,...,a_n\}$ since $X$ is separated, $\{x_i\}$ is closed, so $V=U-\{a_1,...,a_n\}$ is a neighborhood of $x$ such that $V\cap A$ is empty. Contradiction.

On the other hand, suppose that every neighborhood of $x$ contains an infinite number points of $A$, it contains at least a point, by definition, it is an accumulation point.